First, we have
\(3n+1 &= \prod _{p\in \Pi _1}p^{\nu _p(3n+1)} \cdot \prod _{p\in \Pi _2}p^{\nu _p(3n+1)}\\&= \prod _{p\in \Pi _1} x_p^{\nu _p(3n+1)}\, \overline{x}_p^{\nu _p(3n+1)}\cdot \prod _{p\in \Pi _2}p^{\nu _p(3n+1)},\) 

that gives a factorization into irreducible elements in \(\mathcal {R}\) . Now,
we note that there is \(z=\frac{a}{2} + \frac{b}{2} i\sqrt{3} \in \mathcal {R}\) 
such that \(3n+1=\mathcal {N}(z) = z\overline{z}\)  if and only if
\(a^2+3b^2=12n+4\)  if and only if \((a,b)\in \mathcal {U}_{12n+4}\) .
Assume there is \(z\in \mathcal {R}\)  such that \(3n+1=\mathcal {N}(z)\) .
By [1]}, \(\mathcal {R}\)  is a Euclidean
domain. Hence, for any \(p\in \Pi _2\)  such that \(\nu _p(3n+1)\ne 0\) , \(p\) 
divides \(z\)  or \(\overline{z}\)  in \(\mathcal {R}\) . We can assume without loss
of generality that \(p\)  divides \(z\) . Then \(p=\overline{p}\)  divides
\(\overline{z}\) , proving that \(p^2\)  divides \(3n+1\)  in \(\mathbb {Z}\) . It follows that
\(\nu _p(3n+1)\)  is even. Furthermore, by the uniqueness (up to a unit
element) of the decomposition into irreducible elements in a Euclidean
ring, we deduce first that \(q_n\)  divides \(z\) , and then that there are
\(\xi \in \mathcal {R}^{\times }\)  and integers \(0 \le j_p \le \nu _p(3n+1)\) 
and \(0 \le j_p^{\prime } \le \nu _p(3n+1)\)  for any \(p\in \Pi _1\) , such that
\(z=\xi q_n \prod _{p\in \Pi _1} x_p^{j_p} \, {\overline{x_p}}^{j_p^{\prime }}.\) 
Hence,
\(3n+1=z\overline{z}=q_n^2\prod _{p\in \Pi _1} x_p^{j_p+j^{\prime }_p}\ {\overline{x_p}}^{j_p + j_p^{\prime }}.\) 
By the uniqueness of the decomposition, we deduce that \(j_p + j_p^{\prime } =\nu _p(3n+1)\) . It follows that
\(z=\xi q_n \prod _{p\in \Pi _1} x_p^{j_p}\,\overline{x_p}^{\nu _p(3n+1)-j_p} = \xi q_n x_{\underline{j}}=\xi y_{\underline{j}},\) 
with \(\underline{j} = (j_p\mid p\in \Pi _1)\) .
Conversely, any factorization in \(\mathcal {R}\)  of \(3n+1\)  of the form \(z\overline{z}\)  is as in (REF ) for some \(\xi \in \mathcal {R}^{\times }\)  and \(\underline{j} \in \mathcal {V}_n\) .
Now, by [1]}, \(\mathcal {R}^{\times }\) 
has order 6 and \(\mathcal {R}^{\times }=\langle \mathbf {j}+1 \rangle \) . Note
that \(\mathcal {R}^{\times }\)  acts on \(\rho (\mathcal {U}_{12n+4})\)  by
multiplication since the elements of \(\mathcal {R}^{\times }\)  have norm 1.
On the other hand, if \(z = \frac{1}{2}(a+bi\sqrt{3})\)  for \(a,\,b\in \mathbb {Z}\) , then
\((\mathbf {j}+1)z=\frac{1}{2}\left( \frac{1}{2}(a-3b)+\frac{1}{2} (a+b) i \right).\) 
Comparing with (REF ), we see that \(\langle R \rangle \)  acts on
\(\mathcal {C}_n\)  like \(\mathcal {R}^{\times }\)  on \(\rho (\mathcal {U}_{12n+4})\) . In
particular, by Theorem REF , for any \(\underline{j} \in \mathcal {V}_n\) , there is \(\xi \in \mathcal {R}^{\times }\)  such that \(\xi q_n x_{\underline{j}}=\frac{1}{2}(u_{\underline{j}}+v_{\underline{j}} i)\)  satisfies
\(u_{\underline{j}}\equiv v_{\underline{j}}\equiv 1 \mod {3}\) . Then
\(\vartheta (\mathcal {C}_n)=\lbrace (u_{\underline{j}}, v_{\underline{j}}) \mid \underline{j}\in \mathcal {V}_n\rbrace \)  by Theorem REF , as required.
